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JEE • MATHEMATICS • COMPLEX NUMBERS
Complex numbers practice · Medium
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Practice format and chapter quiz
These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.
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Showing 241-270 of 337 questions
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241
Medium
241 of 337
Application
+4 | -1
Let and , then and are roots of the equation.
View answer and explanation
Correct option: A
The correct answer is A () because it is the quadratic equation whose roots are derived from the given complex number equation.
242
Medium
242 of 337
Analyzing
+4 | -1
The imaginary part of zero, if
View answer and explanation
Correct option: C
Option C () is correct because the imaginary part of the expression is zero when .
243
Medium
243 of 337
Application
+4 | -1
The value of is
View answer and explanation
Correct option: D
Option D () is correct because it is the simplified value of the given summation after applying the formula for the sum of a geometric series and using properties of complex numbers.
244
Medium
244 of 337
Application
+4 | -1
If , then is equal to
View answer and explanation
Correct option: C
Option A ((0,2 tan-1 1/4]) is correct because it is the interval for 'a' where the discriminant of the quadratic equation is greater than zero.
245
Medium
245 of 337
Application
+4 | -1
If , then area of the triangle whose vertices are points is
View answer and explanation
Correct option: A
Option A () is correct because the area of the triangle formed by the given vertices is half the square of the magnitude of z.
246
Medium
246 of 337
Application
+4 | -1
If , where and all real numbers, then is
View answer and explanation
Correct option: B
Option B () is correct because the expression simplifies to , and thus u = .
247
Medium
247 of 337
Application
+4 | -1
The centre of a regular hexagon is at the point . If one of its vertices is at , then the adjacent vertices of are at the points
View answer and explanation
Correct option: D
Option D () is correct because it represents the vertices adjacent to in a regular hexagon centered at .
248
Medium
248 of 337
Application
+4 | -1
If and , then is equal to
View answer and explanation
Correct option: A
-2
Option A (-2) is correct because by cubing and equating real and imaginary parts, the expression simplifies to -2.
249
Medium
249 of 337
Application
+4 | -1
If is a variable complex number such that , then
View answer and explanation
Correct option: C
Option C () is correct because by substituting into the given argument equation and simplifying, we obtain the equation of a circle.
250
Medium
250 of 337
Application
+4 | -1
If and are 2 pairs of complex conjugate numbers, then
equals:
equals:
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the arguments of a complex number and its conjugate is zero, and z2 and z4 are conjugates of z1 and z3 respectively.
251
Medium
251 of 337
Application
+4 | -1
If , then is equal to
View answer and explanation
Correct option: B
Option B () is correct because after simplifying the expression using properties of complex numbers, the result matches this option.
252
Medium
252 of 337
Application
+4 | -1
If is a unimodular complex number then
View answer and explanation
Correct option: A
Option A () is correct because after simplification using trigonometric identities, the argument of the expression reduces to half of the argument of z.
253
Medium
253 of 337
Application
+4 | -1
The minimum value of , where and are all not equal integers and is a cube root of unity, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the minimum value of the expression is 1 when a, b, and c are distinct integers.
254
Medium
254 of 337
Application
+4 | -1
Let be three vertices of an equilateral triangle circumscribing the circle If and were in anticlockwise sense, then is
View answer and explanation
Correct option: D
The correct answer is D () because, in an equilateral triangle, the vertices are equally spaced around the circumscribing circle, and using the properties of complex numbers, z2 = -1.
255
Medium
255 of 337
Application
+4 | -1
Principal Argument of complex number is
View answer and explanation
Correct option: C
Option C (-5π/12) is correct because it is the principal argument obtained after calculating the argument of the complex number and adjusting to the range (-π, π].
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256
Medium
256 of 337
Application
+4 | -1
If is equal to
View answer and explanation
Correct option: C
Option C () is correct because it is the simplified form of the given expression using Euler's formula and trigonometric identities.
257
Medium
257 of 337
Application
+4 | -1
Value of is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the summation simplifies to 'i', which is not among the given options.
258
Medium
258 of 337
Application
+4 | -1
The principal value of the and of the complex number are
View answer and explanation
Correct option: A
Option A (-, ) is correct because after simplifying using trigonometric identities and De Moivre's theorem, the argument and modulus are - and , respectively.
259
Medium
259 of 337
Application
+4 | -1
If is a complex number with and , then is a _____ for every .
View answer and explanation
Correct option: B
Rational number
Option B (Rational number) is correct because given the conditions, |z(2n) - 1| can be shown to be a rational number using trigonometric identities and properties of rational numbers.
260
Medium
260 of 337
Application
+4 | -1
If where is a complex constant, then is represented by a point on
View answer and explanation
Correct option: A
A circle
Option A (A circle) is correct because the given equation can be rearranged into the standard form of a circle's equation in the complex plane.
261
Medium
261 of 337
Application
+4 | -1
If , then may be equal to
View answer and explanation
Correct option: C
is multiple of 3
Option C ( is multiple of 3) is correct because when n is a multiple of 3, the expression simplifies to this value.
262
Medium
262 of 337
Application
+4 | -1
The value of
View answer and explanation
Correct option: C
1
Option C (1) is correct because after simplification using Euler's formula, the expression evaluates to 1.
263
Medium
263 of 337
Application
+4 | -1
View answer and explanation
Correct option: C
Option C (-i) is correct because by converting to exponential form and using geometric series formula, the sum simplifies to -i.
264
Medium
264 of 337
Analyzing
+4 | -1
If satisfies the equation , then minimum value of is equal to:
View answer and explanation
Correct option: A
0
The correct answer is A (0) because the given equation simplifies to |z-2| = |z+2|, which represents the perpendicular bisector of the line joining 2 and -2, i.e., the y-axis. The minimum value of |z| is 0, occurring at the origin.
265
Medium
265 of 337
Application
+4 | -1
The multiplicative inverse of is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the multiplicative inverse of is , which is not among the given options.
266
Medium
266 of 337
Application
+4 | -1
If then maximum and minimum value of is .
View answer and explanation
Correct option: C
6,0
Option C (6,0) is correct because the maximum and minimum values of are found by using the triangle inequality on .
267
Medium
267 of 337
Application
+4 | -1
In a geometrical progression, first term and common ratio are both . Then the absolute value of the th term of the progression is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the absolute value of the nth term of the geometric progression is 1, as the magnitude of the common ratio is 1.
268
Medium
268 of 337
Application
+4 | -1
The value of for is
View answer and explanation
Correct option: B
Option B () is correct because substituting into the expression and simplifying results in -160.
269
Medium
269 of 337
Application
+4 | -1
If and , a lower bound of is
View answer and explanation
Correct option: C
Option C () is correct because by using properties of complex numbers and the reverse triangle inequality, we can establish the lower bound for |z|.
270
Medium
270 of 337
Application
+4 | -1
If then maximum and minimum value of is -
View answer and explanation
Correct option: A
8,2
Option A (8,2) is correct because the maximum and minimum values of are found by using the triangle inequality on .
Revision notes and key takeaways
Review COMPLEX NUMBERS in your textbook or class notes, then use this set to check your understanding.
Showing 241-270 of 337 questions
